# Distance traveled from displacement

I am currently reading a book called Physics for Scientists and Engineers by Serway. While reading the chapter about 2-dimensional kinematics, I asked myself a question that I can't solve. The problem in the book is as follows:

A particle is moving in a plane, starting from (0,0). $V_{0x}=20m/s$, $A_x=4m/s$, $V_{0y}=-15m/s.$ Find a,b,c,d.....

From this data I wrote a displacement equation as follow:

$$\vec r = (20t+2(t^2))î - (15t)j$$

Then I asked if it is possible to find the distance traveled at a certain point in time (t=x), and not only the displacement. I couldn't seem to figure out how to do that! Thanks for any help.

Length: $\int_{t_0}^{t_1}| \vec{r} '(t)|dt$. In your case: $\int_{0}^{t}\sqrt{(20+4t')^2+(15)^2}dt'$ Use Wolframalpha if you are now sure how to deal with it: https://www.wolframalpha.com/input/?i=int+sqrt(%2820%2B4x%29^2%2B%2815%29^2)dx
• Exactly. $|\vec r'(t)|dt$ can (and should) be interpreted as the infinitesimal distance travelled between the times $t$ and $t+dt$. Commented Aug 12, 2015 at 19:17
Since this is a specific question, I can't answer it directly for you but I may give you a hint: you already have the displacement formula $\stackrel{\to }{r}\left(t\right)$. From there you may find the relation y(x) using t as a parameter and use the formula for the arc length from calculus: $L=\underset{a}{\overset{b}{\int }}\sqrt{1+{\left({y}^{\prime }\right)}^{2}}dx$. Cheers!